题目
Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example,
Given input array nums &＃61; [1,1,2],
Your function should return length &＃61; 2, with the first two elements of nums being 1 and 2 respectively. It doesn&＃39;t matter what you leave beyond the new length.
题意&＃xff1a;
就是将一个已经排好序的数组进行去重&＃xff0c;并且不考虑数组的去重之后的其他内容。
此题比较简单&＃xff0c;因为之前也有一道去重的题目&＃xff0c;所以做这道题就有点得心应手了。但是这里我用了一个TreeSet来处理&＃xff0c;因为TreeSet会自动进行排序&＃xff0c;一开始我是用HashSet来进行排序的&＃xff0c;后来发现是HashSet是根据HashCode来进行排序的&＃xff0c;所以不能用这个HashSet&＃xff0c;于是我考虑用TreeSet来处理。
public class Solution
{public static int removeDuplicates(int[] nums){int length &＃61; nums.length;if(length &＃61;&＃61; 0)return 0;else if(length &＃61;&＃61; 1)return 1;else{int num &＃61; 0;Set uniqueSet &＃61; new TreeSet();uniqueSet.add(nums[0]);for(int i &＃61; 1; i }
有了上一题做铺垫&＃xff0c;那么接下来的一题就是非常简单的变种&＃xff1a;
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array nums &＃61; [1,1,1,2,2,3],
Your function should return length &＃61; 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn&＃39;t matter what you leave beyond the new length.
只允许在数组中的值出现两次&＃xff0c;然后重新将这个数组进行排列&＃xff0c;并返回删除重复的数字之后的数组的长度&＃xff0c;要求原数组的前n个不重复的数字是新删除之后的。原理同上一题类似&＃xff0c;只不过上一天用hashSet来做&＃xff0c;这一题用HashMap来做。
public int removeDuplicates(int[] nums){int count &＃61; 0;int length &＃61; nums.length;int[] result &＃61; new int[length];HashMap hashMap &＃61; new HashMap();int j &＃61; 0;for(int i &＃61; 0; i
京公网安备 11010802041100号